Question

In a two-slit interference experiment, with a light source of unknown wavelength, the following data are measured:

slit separation, d = 0.22 mm,

distance of slits from the screen, L = 2.39 m, (large enough that the small-angle approximation applies),

separation between the m =

Answer 1

= 532 nm = 532 * 10^-9 m

Relative uncertainty in the measurement of slit separation d = 3.5 %

Thus, absolute uncertainty in the measurement of slit separation, d = (relative uncertainty) * (slit separation)

d = 0.035 * 0.00022 m = 0.0000077 m

Thus, the wavelength is,

= (d d)y_m / m L

= (0.00022 0.0000077) * 0.01735 / (3 * 2.39)

= ( 532 19 ) nm

Thus, absolute uncertainty of wavelength is = 19 nm

Part II

Single slit width w = 0.030 mm = 0.00003 m

Wavelength of the light source = 583 nm = 583 * 10^-9 m

Distance between the screen and the slit is, L = 0.495 m

order m = 4

We know,

w sin = m

sin = 4 * 583 * 10^-9 / 0.00003

sin = 0.077733333 y_m / L

[ is about 4.458 degrees and hence I'm using the distance L for the hypotenuse of the triangle formed by line of incidence, distance of fourth dark fringe from central maximum on the screen and emerged light.]

y_m = 0.077733333 * 0.495 m

y_m = 0.038478 m is the distance of 4th dark fringe from the central bright maximum.